Is my proof of $sqrt2 + sqrt3 + sqrt5$ is an irrational number valid? [duplicate]

Clash Royale CLAN TAG#URR8PPP

up vote
-1
down vote

favorite

2

This question already has an answer here:

• 
  Prove that $sqrt2+sqrt3+sqrt5$ is irrational. Generalise this.
  
  5 answers
  
  

The question is prove $sqrt2 + sqrt3 + sqrt5$ is an irrational number.

I started by assuming the opposite that $sqrt2 + sqrt3 + sqrt5$ is a rational number. I stated that a rational number is a number made by dividing two integers. So I set $sqrt2 + sqrt3 + sqrt5 = i_1/i_2$, where $i_1$ and $i_2$ are two integers. I multiplied $i_2$ onto both sides and got $i_2sqrt2 + i_2sqrt3 + i_2sqrt5 = i_1$. I then said that in order to turn an irrational number such as $sqrt2$ into a rational number you can multiply, $sqrtnsqrtn=n$. Meaning $i_2$ would have to hold the value of $sqrt2$, $sqrt3$ and $sqrt5$ which is impossible. So it is an irrational.

I think I made a mistake somewhere but I am not sure.

proof-verification

share|cite|improve this question

edited Jul 26 at 21:47

rogerl

16.5k22745

asked Jul 26 at 21:43

mathguy21

71

marked as duplicate by Xander Henderson, max_zorn, user21820, Lord Shark the Unknown, Jyrki Lahtonen Jul 27 at 5:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is not clear what does it mean to say that an integer "holds a value"?
  – lulu
  Jul 26 at 21:46
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  in order to turn an irrational number such as sqrt(2) into a rational number you can multiply, sqrt(n)sqrt(n)=n. This is where the mistake is, this really makes no sense. At the very least, it seems to ignore the fact that adding irrational numbers can lead to a rational sum.
  – Arnaud Mortier
  Jul 26 at 21:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The sum of irrational numbers can be rational, even an integer. So , you cannot conclude this way. You need the minimal polynomial and have to show that its degree is larger than $1$ (In this case, it is $8$).
  – Peter
  Jul 26 at 21:46
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  To warm up for this problem, start with something a little simpler like $sqrt 2 +sqrt 3$.
  – lulu
  Jul 26 at 21:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the advice. I will try again and take into consideration what you all mentioned :)
  – mathguy21
  Jul 26 at 21:49
  

  
  
  
  

 | 
show 5 more comments

up vote
-1
down vote

favorite

2

This question already has an answer here:

• 
  Prove that $sqrt2+sqrt3+sqrt5$ is irrational. Generalise this.
  
  5 answers
  
  

The question is prove $sqrt2 + sqrt3 + sqrt5$ is an irrational number.

I started by assuming the opposite that $sqrt2 + sqrt3 + sqrt5$ is a rational number. I stated that a rational number is a number made by dividing two integers. So I set $sqrt2 + sqrt3 + sqrt5 = i_1/i_2$, where $i_1$ and $i_2$ are two integers. I multiplied $i_2$ onto both sides and got $i_2sqrt2 + i_2sqrt3 + i_2sqrt5 = i_1$. I then said that in order to turn an irrational number such as $sqrt2$ into a rational number you can multiply, $sqrtnsqrtn=n$. Meaning $i_2$ would have to hold the value of $sqrt2$, $sqrt3$ and $sqrt5$ which is impossible. So it is an irrational.

I think I made a mistake somewhere but I am not sure.

proof-verification

share|cite|improve this question

edited Jul 26 at 21:47

rogerl

16.5k22745

asked Jul 26 at 21:43

mathguy21

71

marked as duplicate by Xander Henderson, max_zorn, user21820, Lord Shark the Unknown, Jyrki Lahtonen Jul 27 at 5:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is not clear what does it mean to say that an integer "holds a value"?
  – lulu
  Jul 26 at 21:46
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  in order to turn an irrational number such as sqrt(2) into a rational number you can multiply, sqrt(n)sqrt(n)=n. This is where the mistake is, this really makes no sense. At the very least, it seems to ignore the fact that adding irrational numbers can lead to a rational sum.
  – Arnaud Mortier
  Jul 26 at 21:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The sum of irrational numbers can be rational, even an integer. So , you cannot conclude this way. You need the minimal polynomial and have to show that its degree is larger than $1$ (In this case, it is $8$).
  – Peter
  Jul 26 at 21:46
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  To warm up for this problem, start with something a little simpler like $sqrt 2 +sqrt 3$.
  – lulu
  Jul 26 at 21:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the advice. I will try again and take into consideration what you all mentioned :)
  – mathguy21
  Jul 26 at 21:49
  

  
  
  
  

 | 
show 5 more comments

up vote
-1
down vote

favorite

2

up vote
-1
down vote

favorite

2

2

This question already has an answer here:

• 
  Prove that $sqrt2+sqrt3+sqrt5$ is irrational. Generalise this.
  
  5 answers
  
  

The question is prove $sqrt2 + sqrt3 + sqrt5$ is an irrational number.

I started by assuming the opposite that $sqrt2 + sqrt3 + sqrt5$ is a rational number. I stated that a rational number is a number made by dividing two integers. So I set $sqrt2 + sqrt3 + sqrt5 = i_1/i_2$, where $i_1$ and $i_2$ are two integers. I multiplied $i_2$ onto both sides and got $i_2sqrt2 + i_2sqrt3 + i_2sqrt5 = i_1$. I then said that in order to turn an irrational number such as $sqrt2$ into a rational number you can multiply, $sqrtnsqrtn=n$. Meaning $i_2$ would have to hold the value of $sqrt2$, $sqrt3$ and $sqrt5$ which is impossible. So it is an irrational.

I think I made a mistake somewhere but I am not sure.

proof-verification

share|cite|improve this question

edited Jul 26 at 21:47

rogerl

16.5k22745

asked Jul 26 at 21:43

mathguy21

71

This question already has an answer here:

• 
  Prove that $sqrt2+sqrt3+sqrt5$ is irrational. Generalise this.
  
  5 answers
  
  

The question is prove $sqrt2 + sqrt3 + sqrt5$ is an irrational number.

I started by assuming the opposite that $sqrt2 + sqrt3 + sqrt5$ is a rational number. I stated that a rational number is a number made by dividing two integers. So I set $sqrt2 + sqrt3 + sqrt5 = i_1/i_2$, where $i_1$ and $i_2$ are two integers. I multiplied $i_2$ onto both sides and got $i_2sqrt2 + i_2sqrt3 + i_2sqrt5 = i_1$. I then said that in order to turn an irrational number such as $sqrt2$ into a rational number you can multiply, $sqrtnsqrtn=n$. Meaning $i_2$ would have to hold the value of $sqrt2$, $sqrt3$ and $sqrt5$ which is impossible. So it is an irrational.

I think I made a mistake somewhere but I am not sure.

This question already has an answer here:

• 
  Prove that $sqrt2+sqrt3+sqrt5$ is irrational. Generalise this.
  
  5 answers
  
  

proof-verification

share|cite|improve this question

edited Jul 26 at 21:47

rogerl

16.5k22745

asked Jul 26 at 21:43

mathguy21

71

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 21:47

rogerl

16.5k22745

edited Jul 26 at 21:47

rogerl

16.5k22745

edited Jul 26 at 21:47

rogerl

16.5k22745

16.5k22745

asked Jul 26 at 21:43

mathguy21

71

asked Jul 26 at 21:43

mathguy21

71

asked Jul 26 at 21:43

mathguy21

71

71

marked as duplicate by Xander Henderson, max_zorn, user21820, Lord Shark the Unknown, Jyrki Lahtonen Jul 27 at 5:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Xander Henderson, max_zorn, user21820, Lord Shark the Unknown, Jyrki Lahtonen Jul 27 at 5:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is not clear what does it mean to say that an integer "holds a value"?
  – lulu
  Jul 26 at 21:46
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  in order to turn an irrational number such as sqrt(2) into a rational number you can multiply, sqrt(n)sqrt(n)=n. This is where the mistake is, this really makes no sense. At the very least, it seems to ignore the fact that adding irrational numbers can lead to a rational sum.
  – Arnaud Mortier
  Jul 26 at 21:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The sum of irrational numbers can be rational, even an integer. So , you cannot conclude this way. You need the minimal polynomial and have to show that its degree is larger than $1$ (In this case, it is $8$).
  – Peter
  Jul 26 at 21:46
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  To warm up for this problem, start with something a little simpler like $sqrt 2 +sqrt 3$.
  – lulu
  Jul 26 at 21:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the advice. I will try again and take into consideration what you all mentioned :)
  – mathguy21
  Jul 26 at 21:49
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is not clear what does it mean to say that an integer "holds a value"?
  – lulu
  Jul 26 at 21:46
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  in order to turn an irrational number such as sqrt(2) into a rational number you can multiply, sqrt(n)sqrt(n)=n. This is where the mistake is, this really makes no sense. At the very least, it seems to ignore the fact that adding irrational numbers can lead to a rational sum.
  – Arnaud Mortier
  Jul 26 at 21:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The sum of irrational numbers can be rational, even an integer. So , you cannot conclude this way. You need the minimal polynomial and have to show that its degree is larger than $1$ (In this case, it is $8$).
  – Peter
  Jul 26 at 21:46
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  To warm up for this problem, start with something a little simpler like $sqrt 2 +sqrt 3$.
  – lulu
  Jul 26 at 21:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the advice. I will try again and take into consideration what you all mentioned :)
  – mathguy21
  Jul 26 at 21:49
  

  
  
  
  

This is not clear what does it mean to say that an integer "holds a value"?
– lulu
Jul 26 at 21:46

This is not clear what does it mean to say that an integer "holds a value"?
– lulu
Jul 26 at 21:46

2

2

in order to turn an irrational number such as sqrt(2) into a rational number you can multiply, sqrt(n)sqrt(n)=n. This is where the mistake is, this really makes no sense. At the very least, it seems to ignore the fact that adding irrational numbers can lead to a rational sum.
– Arnaud Mortier
Jul 26 at 21:46

in order to turn an irrational number such as sqrt(2) into a rational number you can multiply, sqrt(n)sqrt(n)=n. This is where the mistake is, this really makes no sense. At the very least, it seems to ignore the fact that adding irrational numbers can lead to a rational sum.
– Arnaud Mortier
Jul 26 at 21:46

The sum of irrational numbers can be rational, even an integer. So , you cannot conclude this way. You need the minimal polynomial and have to show that its degree is larger than $1$ (In this case, it is $8$).
– Peter
Jul 26 at 21:46

The sum of irrational numbers can be rational, even an integer. So , you cannot conclude this way. You need the minimal polynomial and have to show that its degree is larger than $1$ (In this case, it is $8$).
– Peter
Jul 26 at 21:46

4

4

To warm up for this problem, start with something a little simpler like $sqrt 2 +sqrt 3$.
– lulu
Jul 26 at 21:47

To warm up for this problem, start with something a little simpler like $sqrt 2 +sqrt 3$.
– lulu
Jul 26 at 21:47

Thank you for the advice. I will try again and take into consideration what you all mentioned :)
– mathguy21
Jul 26 at 21:49

Thank you for the advice. I will try again and take into consideration what you all mentioned :)
– mathguy21
Jul 26 at 21:49

 | 
show 5 more comments

5 Answers
5

active

oldest

votes

up vote
4
down vote

If $sqrt2+sqrt3$ is rational then so too is $sqrt2-sqrt3$ because $(sqrt2+sqrt3)cdot (sqrt2-sqrt3) = 2 - 3 = -1$

But adding the two terms, $(sqrt2+sqrt3)+ (sqrt2-sqrt3) = 2sqrt2$ which is irrational. Two rational numbers cannot sum to an irrational number so we have a contradiction. Therefore $sqrt2+sqrt3$ is irrational. We can say $sqrt2+sqrt3$ = I and come to the same result/conclusion for I$ + sqrt5$.

In this case we reach the assumption that I$^2-5$ is rational.

But I$^2-5= (sqrt2+sqrt3)^2-5 = 5+2sqrt6-5 = 2sqrt6$ which is irrational and another contradiction. Hence $sqrt2+sqrt3+sqrt5$ is irrational.

share|cite|improve this answer

edited Jul 26 at 23:25

answered Jul 26 at 22:36

Phil H

1,8132311

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How do you do the proof for $I+sqrt5$ being irrational? One key property of the original proof would require$(I+sqrt5)(I-sqrt5)$ being rational, but it equals the irrational $I^2-5=2+3+2sqrt6-5 = 2sqrt6$. OK, good argument!
  – Ingix
  Jul 26 at 22:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ Ingix We were working on it at the same time apparently,
  – Phil H
  Jul 26 at 22:59
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I'd suggest you correct your final evaluation of $I^2$. It doesn't change the outcome of the argument, but is confusing for somebody follwing the proof.
  – Ingix
  Jul 26 at 23:02
  

  
  
  
  

add a comment | 

up vote
2
down vote

No, this solution is incorrect I'm afraid. You need $i_2sqrt2+i_2sqrt3+i_2sqrt5=i_1$, that's true, but that doesn't mean that $i_2sqrt2$ needs to be rational. Also, if it did need to be rational, that doesn't mean that $i_2$ must be $sqrt2$. In can also be $3sqrt2/5$, for example.

share|cite|improve this answer

answered Jul 26 at 22:05

Alon Amit

10.2k3765

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the clarification!
  – mathguy21
  Jul 26 at 22:12
  

  
  
  
  

add a comment | 

up vote
0
down vote

Let $x=sqrt 2+sqrt 3+sqrt 5$ and $y=x-sqrt 5=sqrt 2+sqrt 3.$. We have $$0=(y-sqrt 2-sqrt 3)(y-sqrt 2+sqrt 3)=(y-sqrt 2)^2-(sqrt 3)^2=y^2-2ysqrt 2-1.$$
So $y^2-1=2ysqrt 2.$ Squaring this and re-grouping, we have $y^4-10y^2+1=0.$ Substituting $x-sqrt 5$ for $y$ in this, expanding and re-grouping, we obtain $$A=Bsqrt 5 quad text wherequad A=x^4+4x^2-24quad text andquad B=20x^3.$$

Now if $x$ is rational then $A$ and $B$ are rational with $B=20x^3ne 0, $ implying $sqrt 5=A/Bin Bbb Q, $ which is false. So $x$ cannot be rational.

share|cite|improve this answer

answered Jul 26 at 22:56

DanielWainfleet

31.5k31542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  We could go further to obtain a polynomial $p(z)in Bbb Z[z]$ with deg $(p)=8$ and $p(x)=0$ and then prove that $p(z)$ is irreducible in $Bbb Z[z]$ so, by a theorem of Gauss, $p(z)$ is irreducible in $Bbb Q[z]$. But it suffices to stop at this intermediate stage.
  – DanielWainfleet
  Jul 30 at 19:10
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

Consider $r$ rational and assume

$$sqrt 2 + sqrt 3 + sqrt 5=r $$

then

$$iff sqrt 2 + sqrt 3)=r - sqrt 5$$

$$5+ 2sqrt 6=r^2 + 5-2r sqrt 5$$

$$2sqrt 6 + 2r sqrt 5=r^2$$

$$24+20r^2+8r sqrt30=r^4$$

$$8r sqrt30=r^4-24-20r^2+$$

$$sqrt30=fracr^38-frac 3 r-frac52 r$$

then it suffices to show that $sqrt 30$ is not rational (which is true since $30$ is not a perfect square).

share|cite|improve this answer

edited Jul 27 at 14:36

answered Jul 26 at 22:05

gimusi

65k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In the last line what property is being shown that shows root(30) is irrational.?
  – mathguy21
  Jul 26 at 22:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathguy21 30 is not a perfect square.
  – Phil H
  Jul 26 at 22:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathguy21 We are using contradiction and the initial assumption is true only if the last one is true, therfore we need to prove that $sqrt 30$ is irrational.
  – gimusi
  Jul 26 at 22:15
  

  
  
  
  

add a comment | 

up vote
-2
down vote

The sum of two irrational numbers can be rational, so you cannot go from your hypothesis that $i_2sqrt2 + i_2sqrt 3+i_2sqrt 5$ is rational to the claim that each of them is separately rational. An example would be $3-sqrt 2$ and $sqrt 2$ are both irrational but their sum is $3$.

For the sum of two square roots, like $sqrt 2 + sqrt 3$ you can use the fact that the square of a rational is rational. You can assume it is rational and square it to get $5+2sqrt 6$. As you are down to one square root you can follow the usual proof that $sqrt 2$ is irrational.

share|cite|improve this answer

edited Jul 26 at 23:10

Xander Henderson

13.1k83150

answered Jul 26 at 22:06

Ross Millikan

275k21186351

add a comment | 

5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

If $sqrt2+sqrt3$ is rational then so too is $sqrt2-sqrt3$ because $(sqrt2+sqrt3)cdot (sqrt2-sqrt3) = 2 - 3 = -1$

But adding the two terms, $(sqrt2+sqrt3)+ (sqrt2-sqrt3) = 2sqrt2$ which is irrational. Two rational numbers cannot sum to an irrational number so we have a contradiction. Therefore $sqrt2+sqrt3$ is irrational. We can say $sqrt2+sqrt3$ = I and come to the same result/conclusion for I$ + sqrt5$.

In this case we reach the assumption that I$^2-5$ is rational.

But I$^2-5= (sqrt2+sqrt3)^2-5 = 5+2sqrt6-5 = 2sqrt6$ which is irrational and another contradiction. Hence $sqrt2+sqrt3+sqrt5$ is irrational.

share|cite|improve this answer

edited Jul 26 at 23:25

answered Jul 26 at 22:36

Phil H

1,8132311

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How do you do the proof for $I+sqrt5$ being irrational? One key property of the original proof would require$(I+sqrt5)(I-sqrt5)$ being rational, but it equals the irrational $I^2-5=2+3+2sqrt6-5 = 2sqrt6$. OK, good argument!
  – Ingix
  Jul 26 at 22:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ Ingix We were working on it at the same time apparently,
  – Phil H
  Jul 26 at 22:59
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I'd suggest you correct your final evaluation of $I^2$. It doesn't change the outcome of the argument, but is confusing for somebody follwing the proof.
  – Ingix
  Jul 26 at 23:02
  

  
  
  
  

add a comment | 

up vote
4
down vote

If $sqrt2+sqrt3$ is rational then so too is $sqrt2-sqrt3$ because $(sqrt2+sqrt3)cdot (sqrt2-sqrt3) = 2 - 3 = -1$

But adding the two terms, $(sqrt2+sqrt3)+ (sqrt2-sqrt3) = 2sqrt2$ which is irrational. Two rational numbers cannot sum to an irrational number so we have a contradiction. Therefore $sqrt2+sqrt3$ is irrational. We can say $sqrt2+sqrt3$ = I and come to the same result/conclusion for I$ + sqrt5$.

In this case we reach the assumption that I$^2-5$ is rational.

But I$^2-5= (sqrt2+sqrt3)^2-5 = 5+2sqrt6-5 = 2sqrt6$ which is irrational and another contradiction. Hence $sqrt2+sqrt3+sqrt5$ is irrational.

share|cite|improve this answer

edited Jul 26 at 23:25

answered Jul 26 at 22:36

Phil H

1,8132311

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How do you do the proof for $I+sqrt5$ being irrational? One key property of the original proof would require$(I+sqrt5)(I-sqrt5)$ being rational, but it equals the irrational $I^2-5=2+3+2sqrt6-5 = 2sqrt6$. OK, good argument!
  – Ingix
  Jul 26 at 22:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ Ingix We were working on it at the same time apparently,
  – Phil H
  Jul 26 at 22:59
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I'd suggest you correct your final evaluation of $I^2$. It doesn't change the outcome of the argument, but is confusing for somebody follwing the proof.
  – Ingix
  Jul 26 at 23:02
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

If $sqrt2+sqrt3$ is rational then so too is $sqrt2-sqrt3$ because $(sqrt2+sqrt3)cdot (sqrt2-sqrt3) = 2 - 3 = -1$

But adding the two terms, $(sqrt2+sqrt3)+ (sqrt2-sqrt3) = 2sqrt2$ which is irrational. Two rational numbers cannot sum to an irrational number so we have a contradiction. Therefore $sqrt2+sqrt3$ is irrational. We can say $sqrt2+sqrt3$ = I and come to the same result/conclusion for I$ + sqrt5$.

In this case we reach the assumption that I$^2-5$ is rational.

But I$^2-5= (sqrt2+sqrt3)^2-5 = 5+2sqrt6-5 = 2sqrt6$ which is irrational and another contradiction. Hence $sqrt2+sqrt3+sqrt5$ is irrational.

share|cite|improve this answer

edited Jul 26 at 23:25

answered Jul 26 at 22:36

Phil H

1,8132311

If $sqrt2+sqrt3$ is rational then so too is $sqrt2-sqrt3$ because $(sqrt2+sqrt3)cdot (sqrt2-sqrt3) = 2 - 3 = -1$

But adding the two terms, $(sqrt2+sqrt3)+ (sqrt2-sqrt3) = 2sqrt2$ which is irrational. Two rational numbers cannot sum to an irrational number so we have a contradiction. Therefore $sqrt2+sqrt3$ is irrational. We can say $sqrt2+sqrt3$ = I and come to the same result/conclusion for I$ + sqrt5$.

In this case we reach the assumption that I$^2-5$ is rational.

But I$^2-5= (sqrt2+sqrt3)^2-5 = 5+2sqrt6-5 = 2sqrt6$ which is irrational and another contradiction. Hence $sqrt2+sqrt3+sqrt5$ is irrational.

share|cite|improve this answer

edited Jul 26 at 23:25

answered Jul 26 at 22:36

Phil H

1,8132311

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 23:25

edited Jul 26 at 23:25

edited Jul 26 at 23:25

answered Jul 26 at 22:36

Phil H

1,8132311

answered Jul 26 at 22:36

Phil H

1,8132311

answered Jul 26 at 22:36

Phil H

1,8132311

1,8132311

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How do you do the proof for $I+sqrt5$ being irrational? One key property of the original proof would require$(I+sqrt5)(I-sqrt5)$ being rational, but it equals the irrational $I^2-5=2+3+2sqrt6-5 = 2sqrt6$. OK, good argument!
  – Ingix
  Jul 26 at 22:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ Ingix We were working on it at the same time apparently,
  – Phil H
  Jul 26 at 22:59
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I'd suggest you correct your final evaluation of $I^2$. It doesn't change the outcome of the argument, but is confusing for somebody follwing the proof.
  – Ingix
  Jul 26 at 23:02
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How do you do the proof for $I+sqrt5$ being irrational? One key property of the original proof would require$(I+sqrt5)(I-sqrt5)$ being rational, but it equals the irrational $I^2-5=2+3+2sqrt6-5 = 2sqrt6$. OK, good argument!
  – Ingix
  Jul 26 at 22:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ Ingix We were working on it at the same time apparently,
  – Phil H
  Jul 26 at 22:59
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I'd suggest you correct your final evaluation of $I^2$. It doesn't change the outcome of the argument, but is confusing for somebody follwing the proof.
  – Ingix
  Jul 26 at 23:02
  

  
  
  
  

1

1

How do you do the proof for $I+sqrt5$ being irrational? One key property of the original proof would require$(I+sqrt5)(I-sqrt5)$ being rational, but it equals the irrational $I^2-5=2+3+2sqrt6-5 = 2sqrt6$. OK, good argument!
– Ingix
Jul 26 at 22:57

How do you do the proof for $I+sqrt5$ being irrational? One key property of the original proof would require$(I+sqrt5)(I-sqrt5)$ being rational, but it equals the irrational $I^2-5=2+3+2sqrt6-5 = 2sqrt6$. OK, good argument!
– Ingix
Jul 26 at 22:57

@ Ingix We were working on it at the same time apparently,
– Phil H
Jul 26 at 22:59

@ Ingix We were working on it at the same time apparently,
– Phil H
Jul 26 at 22:59

2

2

I'd suggest you correct your final evaluation of $I^2$. It doesn't change the outcome of the argument, but is confusing for somebody follwing the proof.
– Ingix
Jul 26 at 23:02

I'd suggest you correct your final evaluation of $I^2$. It doesn't change the outcome of the argument, but is confusing for somebody follwing the proof.
– Ingix
Jul 26 at 23:02

add a comment | 

up vote
2
down vote

No, this solution is incorrect I'm afraid. You need $i_2sqrt2+i_2sqrt3+i_2sqrt5=i_1$, that's true, but that doesn't mean that $i_2sqrt2$ needs to be rational. Also, if it did need to be rational, that doesn't mean that $i_2$ must be $sqrt2$. In can also be $3sqrt2/5$, for example.

share|cite|improve this answer

answered Jul 26 at 22:05

Alon Amit

10.2k3765

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the clarification!
  – mathguy21
  Jul 26 at 22:12
  

  
  
  
  

add a comment | 

up vote
2
down vote

No, this solution is incorrect I'm afraid. You need $i_2sqrt2+i_2sqrt3+i_2sqrt5=i_1$, that's true, but that doesn't mean that $i_2sqrt2$ needs to be rational. Also, if it did need to be rational, that doesn't mean that $i_2$ must be $sqrt2$. In can also be $3sqrt2/5$, for example.

share|cite|improve this answer

answered Jul 26 at 22:05

Alon Amit

10.2k3765

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the clarification!
  – mathguy21
  Jul 26 at 22:12
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

No, this solution is incorrect I'm afraid. You need $i_2sqrt2+i_2sqrt3+i_2sqrt5=i_1$, that's true, but that doesn't mean that $i_2sqrt2$ needs to be rational. Also, if it did need to be rational, that doesn't mean that $i_2$ must be $sqrt2$. In can also be $3sqrt2/5$, for example.

share|cite|improve this answer

answered Jul 26 at 22:05

Alon Amit

10.2k3765

No, this solution is incorrect I'm afraid. You need $i_2sqrt2+i_2sqrt3+i_2sqrt5=i_1$, that's true, but that doesn't mean that $i_2sqrt2$ needs to be rational. Also, if it did need to be rational, that doesn't mean that $i_2$ must be $sqrt2$. In can also be $3sqrt2/5$, for example.

share|cite|improve this answer

answered Jul 26 at 22:05

Alon Amit

10.2k3765

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 22:05

Alon Amit

10.2k3765

answered Jul 26 at 22:05

Alon Amit

10.2k3765

answered Jul 26 at 22:05

Alon Amit

10.2k3765

10.2k3765

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the clarification!
  – mathguy21
  Jul 26 at 22:12
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the clarification!
  – mathguy21
  Jul 26 at 22:12
  

  
  
  
  

Thanks for the clarification!
– mathguy21
Jul 26 at 22:12

Thanks for the clarification!
– mathguy21
Jul 26 at 22:12

add a comment | 

up vote
0
down vote

Let $x=sqrt 2+sqrt 3+sqrt 5$ and $y=x-sqrt 5=sqrt 2+sqrt 3.$. We have $$0=(y-sqrt 2-sqrt 3)(y-sqrt 2+sqrt 3)=(y-sqrt 2)^2-(sqrt 3)^2=y^2-2ysqrt 2-1.$$
So $y^2-1=2ysqrt 2.$ Squaring this and re-grouping, we have $y^4-10y^2+1=0.$ Substituting $x-sqrt 5$ for $y$ in this, expanding and re-grouping, we obtain $$A=Bsqrt 5 quad text wherequad A=x^4+4x^2-24quad text andquad B=20x^3.$$

Now if $x$ is rational then $A$ and $B$ are rational with $B=20x^3ne 0, $ implying $sqrt 5=A/Bin Bbb Q, $ which is false. So $x$ cannot be rational.

share|cite|improve this answer

answered Jul 26 at 22:56

DanielWainfleet

31.5k31542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  We could go further to obtain a polynomial $p(z)in Bbb Z[z]$ with deg $(p)=8$ and $p(x)=0$ and then prove that $p(z)$ is irreducible in $Bbb Z[z]$ so, by a theorem of Gauss, $p(z)$ is irreducible in $Bbb Q[z]$. But it suffices to stop at this intermediate stage.
  – DanielWainfleet
  Jul 30 at 19:10
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

Let $x=sqrt 2+sqrt 3+sqrt 5$ and $y=x-sqrt 5=sqrt 2+sqrt 3.$. We have $$0=(y-sqrt 2-sqrt 3)(y-sqrt 2+sqrt 3)=(y-sqrt 2)^2-(sqrt 3)^2=y^2-2ysqrt 2-1.$$
So $y^2-1=2ysqrt 2.$ Squaring this and re-grouping, we have $y^4-10y^2+1=0.$ Substituting $x-sqrt 5$ for $y$ in this, expanding and re-grouping, we obtain $$A=Bsqrt 5 quad text wherequad A=x^4+4x^2-24quad text andquad B=20x^3.$$

Now if $x$ is rational then $A$ and $B$ are rational with $B=20x^3ne 0, $ implying $sqrt 5=A/Bin Bbb Q, $ which is false. So $x$ cannot be rational.

share|cite|improve this answer

answered Jul 26 at 22:56

DanielWainfleet

31.5k31542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  We could go further to obtain a polynomial $p(z)in Bbb Z[z]$ with deg $(p)=8$ and $p(x)=0$ and then prove that $p(z)$ is irreducible in $Bbb Z[z]$ so, by a theorem of Gauss, $p(z)$ is irreducible in $Bbb Q[z]$. But it suffices to stop at this intermediate stage.
  – DanielWainfleet
  Jul 30 at 19:10
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let $x=sqrt 2+sqrt 3+sqrt 5$ and $y=x-sqrt 5=sqrt 2+sqrt 3.$. We have $$0=(y-sqrt 2-sqrt 3)(y-sqrt 2+sqrt 3)=(y-sqrt 2)^2-(sqrt 3)^2=y^2-2ysqrt 2-1.$$
So $y^2-1=2ysqrt 2.$ Squaring this and re-grouping, we have $y^4-10y^2+1=0.$ Substituting $x-sqrt 5$ for $y$ in this, expanding and re-grouping, we obtain $$A=Bsqrt 5 quad text wherequad A=x^4+4x^2-24quad text andquad B=20x^3.$$

Now if $x$ is rational then $A$ and $B$ are rational with $B=20x^3ne 0, $ implying $sqrt 5=A/Bin Bbb Q, $ which is false. So $x$ cannot be rational.

share|cite|improve this answer

answered Jul 26 at 22:56

DanielWainfleet

31.5k31542

Let $x=sqrt 2+sqrt 3+sqrt 5$ and $y=x-sqrt 5=sqrt 2+sqrt 3.$. We have $$0=(y-sqrt 2-sqrt 3)(y-sqrt 2+sqrt 3)=(y-sqrt 2)^2-(sqrt 3)^2=y^2-2ysqrt 2-1.$$
So $y^2-1=2ysqrt 2.$ Squaring this and re-grouping, we have $y^4-10y^2+1=0.$ Substituting $x-sqrt 5$ for $y$ in this, expanding and re-grouping, we obtain $$A=Bsqrt 5 quad text wherequad A=x^4+4x^2-24quad text andquad B=20x^3.$$

Now if $x$ is rational then $A$ and $B$ are rational with $B=20x^3ne 0, $ implying $sqrt 5=A/Bin Bbb Q, $ which is false. So $x$ cannot be rational.

share|cite|improve this answer

answered Jul 26 at 22:56

DanielWainfleet

31.5k31542

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 22:56

DanielWainfleet

31.5k31542

answered Jul 26 at 22:56

DanielWainfleet

31.5k31542

answered Jul 26 at 22:56

DanielWainfleet

31.5k31542

31.5k31542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  We could go further to obtain a polynomial $p(z)in Bbb Z[z]$ with deg $(p)=8$ and $p(x)=0$ and then prove that $p(z)$ is irreducible in $Bbb Z[z]$ so, by a theorem of Gauss, $p(z)$ is irreducible in $Bbb Q[z]$. But it suffices to stop at this intermediate stage.
  – DanielWainfleet
  Jul 30 at 19:10
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  We could go further to obtain a polynomial $p(z)in Bbb Z[z]$ with deg $(p)=8$ and $p(x)=0$ and then prove that $p(z)$ is irreducible in $Bbb Z[z]$ so, by a theorem of Gauss, $p(z)$ is irreducible in $Bbb Q[z]$. But it suffices to stop at this intermediate stage.
  – DanielWainfleet
  Jul 30 at 19:10
  
  

  
  
  
  

We could go further to obtain a polynomial $p(z)in Bbb Z[z]$ with deg $(p)=8$ and $p(x)=0$ and then prove that $p(z)$ is irreducible in $Bbb Z[z]$ so, by a theorem of Gauss, $p(z)$ is irreducible in $Bbb Q[z]$. But it suffices to stop at this intermediate stage.
– DanielWainfleet
Jul 30 at 19:10

We could go further to obtain a polynomial $p(z)in Bbb Z[z]$ with deg $(p)=8$ and $p(x)=0$ and then prove that $p(z)$ is irreducible in $Bbb Z[z]$ so, by a theorem of Gauss, $p(z)$ is irreducible in $Bbb Q[z]$. But it suffices to stop at this intermediate stage.
– DanielWainfleet
Jul 30 at 19:10

add a comment | 

up vote
0
down vote

Consider $r$ rational and assume

$$sqrt 2 + sqrt 3 + sqrt 5=r $$

then

$$iff sqrt 2 + sqrt 3)=r - sqrt 5$$

$$5+ 2sqrt 6=r^2 + 5-2r sqrt 5$$

$$2sqrt 6 + 2r sqrt 5=r^2$$

$$24+20r^2+8r sqrt30=r^4$$

$$8r sqrt30=r^4-24-20r^2+$$

$$sqrt30=fracr^38-frac 3 r-frac52 r$$

then it suffices to show that $sqrt 30$ is not rational (which is true since $30$ is not a perfect square).

share|cite|improve this answer

edited Jul 27 at 14:36

answered Jul 26 at 22:05

gimusi

65k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In the last line what property is being shown that shows root(30) is irrational.?
  – mathguy21
  Jul 26 at 22:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathguy21 30 is not a perfect square.
  – Phil H
  Jul 26 at 22:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathguy21 We are using contradiction and the initial assumption is true only if the last one is true, therfore we need to prove that $sqrt 30$ is irrational.
  – gimusi
  Jul 26 at 22:15
  

  
  
  
  

add a comment | 

up vote
0
down vote

Consider $r$ rational and assume

$$sqrt 2 + sqrt 3 + sqrt 5=r $$

then

$$iff sqrt 2 + sqrt 3)=r - sqrt 5$$

$$5+ 2sqrt 6=r^2 + 5-2r sqrt 5$$

$$2sqrt 6 + 2r sqrt 5=r^2$$

$$24+20r^2+8r sqrt30=r^4$$

$$8r sqrt30=r^4-24-20r^2+$$

$$sqrt30=fracr^38-frac 3 r-frac52 r$$

then it suffices to show that $sqrt 30$ is not rational (which is true since $30$ is not a perfect square).

share|cite|improve this answer

edited Jul 27 at 14:36

answered Jul 26 at 22:05

gimusi

65k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In the last line what property is being shown that shows root(30) is irrational.?
  – mathguy21
  Jul 26 at 22:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathguy21 30 is not a perfect square.
  – Phil H
  Jul 26 at 22:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathguy21 We are using contradiction and the initial assumption is true only if the last one is true, therfore we need to prove that $sqrt 30$ is irrational.
  – gimusi
  Jul 26 at 22:15
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Consider $r$ rational and assume

$$sqrt 2 + sqrt 3 + sqrt 5=r $$

then

$$iff sqrt 2 + sqrt 3)=r - sqrt 5$$

$$5+ 2sqrt 6=r^2 + 5-2r sqrt 5$$

$$2sqrt 6 + 2r sqrt 5=r^2$$

$$24+20r^2+8r sqrt30=r^4$$

$$8r sqrt30=r^4-24-20r^2+$$

$$sqrt30=fracr^38-frac 3 r-frac52 r$$

then it suffices to show that $sqrt 30$ is not rational (which is true since $30$ is not a perfect square).

share|cite|improve this answer

edited Jul 27 at 14:36

answered Jul 26 at 22:05

gimusi

65k73583

Consider $r$ rational and assume

$$sqrt 2 + sqrt 3 + sqrt 5=r $$

then

$$iff sqrt 2 + sqrt 3)=r - sqrt 5$$

$$5+ 2sqrt 6=r^2 + 5-2r sqrt 5$$

$$2sqrt 6 + 2r sqrt 5=r^2$$

$$24+20r^2+8r sqrt30=r^4$$

$$8r sqrt30=r^4-24-20r^2+$$

$$sqrt30=fracr^38-frac 3 r-frac52 r$$

then it suffices to show that $sqrt 30$ is not rational (which is true since $30$ is not a perfect square).

share|cite|improve this answer

edited Jul 27 at 14:36

answered Jul 26 at 22:05

gimusi

65k73583

share|cite|improve this answer

share|cite|improve this answer

edited Jul 27 at 14:36

edited Jul 27 at 14:36

edited Jul 27 at 14:36

answered Jul 26 at 22:05

gimusi

65k73583

answered Jul 26 at 22:05

gimusi

65k73583

answered Jul 26 at 22:05

gimusi

65k73583

65k73583

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In the last line what property is being shown that shows root(30) is irrational.?
  – mathguy21
  Jul 26 at 22:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathguy21 30 is not a perfect square.
  – Phil H
  Jul 26 at 22:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathguy21 We are using contradiction and the initial assumption is true only if the last one is true, therfore we need to prove that $sqrt 30$ is irrational.
  – gimusi
  Jul 26 at 22:15
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In the last line what property is being shown that shows root(30) is irrational.?
  – mathguy21
  Jul 26 at 22:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathguy21 30 is not a perfect square.
  – Phil H
  Jul 26 at 22:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathguy21 We are using contradiction and the initial assumption is true only if the last one is true, therfore we need to prove that $sqrt 30$ is irrational.
  – gimusi
  Jul 26 at 22:15
  

  
  
  
  

In the last line what property is being shown that shows root(30) is irrational.?
– mathguy21
Jul 26 at 22:12

In the last line what property is being shown that shows root(30) is irrational.?
– mathguy21
Jul 26 at 22:12

@mathguy21 30 is not a perfect square.
– Phil H
Jul 26 at 22:14

@mathguy21 30 is not a perfect square.
– Phil H
Jul 26 at 22:14

@mathguy21 We are using contradiction and the initial assumption is true only if the last one is true, therfore we need to prove that $sqrt 30$ is irrational.
– gimusi
Jul 26 at 22:15

@mathguy21 We are using contradiction and the initial assumption is true only if the last one is true, therfore we need to prove that $sqrt 30$ is irrational.
– gimusi
Jul 26 at 22:15

add a comment | 

up vote
-2
down vote

The sum of two irrational numbers can be rational, so you cannot go from your hypothesis that $i_2sqrt2 + i_2sqrt 3+i_2sqrt 5$ is rational to the claim that each of them is separately rational. An example would be $3-sqrt 2$ and $sqrt 2$ are both irrational but their sum is $3$.

For the sum of two square roots, like $sqrt 2 + sqrt 3$ you can use the fact that the square of a rational is rational. You can assume it is rational and square it to get $5+2sqrt 6$. As you are down to one square root you can follow the usual proof that $sqrt 2$ is irrational.

share|cite|improve this answer

edited Jul 26 at 23:10

Xander Henderson

13.1k83150

answered Jul 26 at 22:06

Ross Millikan

275k21186351

add a comment | 

up vote
-2
down vote

The sum of two irrational numbers can be rational, so you cannot go from your hypothesis that $i_2sqrt2 + i_2sqrt 3+i_2sqrt 5$ is rational to the claim that each of them is separately rational. An example would be $3-sqrt 2$ and $sqrt 2$ are both irrational but their sum is $3$.

For the sum of two square roots, like $sqrt 2 + sqrt 3$ you can use the fact that the square of a rational is rational. You can assume it is rational and square it to get $5+2sqrt 6$. As you are down to one square root you can follow the usual proof that $sqrt 2$ is irrational.

share|cite|improve this answer

edited Jul 26 at 23:10

Xander Henderson

13.1k83150

answered Jul 26 at 22:06

Ross Millikan

275k21186351

add a comment | 

up vote
-2
down vote

up vote
-2
down vote

The sum of two irrational numbers can be rational, so you cannot go from your hypothesis that $i_2sqrt2 + i_2sqrt 3+i_2sqrt 5$ is rational to the claim that each of them is separately rational. An example would be $3-sqrt 2$ and $sqrt 2$ are both irrational but their sum is $3$.

For the sum of two square roots, like $sqrt 2 + sqrt 3$ you can use the fact that the square of a rational is rational. You can assume it is rational and square it to get $5+2sqrt 6$. As you are down to one square root you can follow the usual proof that $sqrt 2$ is irrational.

share|cite|improve this answer

edited Jul 26 at 23:10

Xander Henderson

13.1k83150

answered Jul 26 at 22:06

Ross Millikan

275k21186351

The sum of two irrational numbers can be rational, so you cannot go from your hypothesis that $i_2sqrt2 + i_2sqrt 3+i_2sqrt 5$ is rational to the claim that each of them is separately rational. An example would be $3-sqrt 2$ and $sqrt 2$ are both irrational but their sum is $3$.

For the sum of two square roots, like $sqrt 2 + sqrt 3$ you can use the fact that the square of a rational is rational. You can assume it is rational and square it to get $5+2sqrt 6$. As you are down to one square root you can follow the usual proof that $sqrt 2$ is irrational.

share|cite|improve this answer

edited Jul 26 at 23:10

Xander Henderson

13.1k83150

answered Jul 26 at 22:06

Ross Millikan

275k21186351

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 23:10

Xander Henderson

13.1k83150

edited Jul 26 at 23:10

Xander Henderson

13.1k83150

edited Jul 26 at 23:10

Xander Henderson

13.1k83150

13.1k83150

answered Jul 26 at 22:06

Ross Millikan

275k21186351

answered Jul 26 at 22:06

Ross Millikan

275k21186351

answered Jul 26 at 22:06

Ross Millikan

275k21186351

275k21186351

add a comment | 

add a comment |